Similarity only

We note that $\triangle DEF$ and $\triangle XYZ$ are similar. As area of similar triangle is directly proportional to square of its side length, if the side lengths of $\triangle DEF$ are $a$ , $b$ , and $c$ then the respective side lengths of $\triangle XYZ$ are $\sqrt 7 a$ , $\sqrt 7 b$ , and $\sqrt 7 c$ . We also note that $\triangle XYZ$ is compose of three parallelograms and $\triangle DEF$ . Then we have:

$\begin{aligned} [YZFE] + [ZXDF] + [XYED] + [DEF] & = [XYZ] \\ \left(\frac {a+\sqrt 7 a}2 \right) \blue{h_a} + \left(\frac {b+\sqrt 7 b}2 \right) \blue{h_b} + \left(\frac {c+\sqrt 7 c}2 \right) \blue{h_c} + 6 & = 42 & \small \blue{\text{where }h_x \text{ is the height of parallelogram.}} \end{aligned}$

$\implies ah_a + bh_b + ch_c = \dfrac {2(42-6)}{1+\sqrt 7} = 12 \left(\sqrt 7 -1\right)$

And we have:

$\begin{aligned} [ABC] & = [AEF] + [BDF] + [CDE] + [DEF] \\ & = \frac 12 \left(ah_a + bh_b + ch_c\right) + 6 \\ & = 6 \left(\sqrt 7 -1\right) - 6 \\ & = 6\sqrt 7 = \sqrt{252} \end{aligned}$

Therefore $k = \boxed{252}$ .